Minimalist IAS
CSAT

CSAT · 43 questions

Counting, permutations & probability

Every UPSC CSAT question on this topic, 2016–2026, newest first. Tap an option to check yourself; the answer and explanation open below it.

Counting, permutations & probability questions per year: 2016: 4, 2017: 3, 2018: 4, 2019: 4, 2020: 2, 2021: 3, 2022: 8, 2023: 11, 2024: 1, 2025: 0, 2026: 3 Asked in 10 of 11 years · most in 2023 (11)

UPSC syllabus: “Basic numeracy (numbers and their relations, orders of magnitude, etc.) (Class X level), Data interpretation (charts, graphs, tables, data sufficiency etc. — Class X level);” See the full syllabus →

CSAT 2022 · Q17

Easy

In a tournament of Chess having 150 entrants, a player is eliminated whenever he loses a match. It is given that no match results in a tie/draw. How many matches are played in the entire tournament?

Answer & explanation

Answer: (c) 149

Every match has exactly one loser, and every loser is eliminated. The tournament ends when one player is left, so 149 players are eliminated — and that takes exactly 149 matches.

  1. Each match produces exactly one loser, who is eliminated.
  2. To leave a single winner, 150 − 1 = 149 players must be eliminated.
  3. So 149 matches are played.
  4. Check: with 4 players, 2 semi-finals + 1 final = 3 = 4 − 1 matches.

Remember · Knockout tournament with n players: always n − 1 matches, whatever the byes.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2022 · Q18

Easy

How many 3-digit natural numbers (without repetition of digits) are there such that each digit is odd and the number is divisible by 5?

Answer & explanation

Answer: (b) 12

A number divisible by 5 with an odd units digit must end in 5. The other two places are filled from 1, 3, 7, 9 without repetition: 4 × 3 = 12 numbers.

  1. Divisible by 5 means the units digit is 0 or 5; it must be odd, so it is 5.
  2. The hundreds digit can be any of the remaining odd digits 1, 3, 7, 9: 4 choices.
  3. The tens digit can be any of the 3 odd digits still unused: 3 choices.
  4. Total = 4 × 3 = 12.

Remember · Fill the restricted place first (here the units digit), then count the choices for the free places.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2022 · Q24

Easy

The letters A, B, C, D and E are arranged in such a way that there are exactly two letters between A and E. How many such arrangements are possible?

Answer & explanation

Answer: (c) 24

In a row of five, A and E can stand only at places 1 and 4 or 2 and 5, in either order: 4 ways. The other three letters then fill the remaining places in 3! = 6 ways, giving 24.

  1. Number the places 1 to 5. Exactly two letters between A and E means A and E occupy places (1, 4) or (2, 5): 2 choices of places.
  2. A and E can swap within the chosen places: 2 × 2 = 4 ways.
  3. B, C and D fill the other three places in 3! = 6 ways.
  4. Total = 4 × 6 = 24.

Remember · Place the constrained pair first (positions × internal order), then arrange the remaining items freely.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2022 · Q27

Easy

A, B and C are three places such that there are three different roads from A to B, four different roads from B to C and three different roads from A to C. In how many different ways can one travel from A to C using these roads?

Answer & explanation

Answer: (c) 15

One can go directly from A to C (3 ways) or go through B (3 × 4 = 12 ways). These are alternatives, so they add: 3 + 12 = 15.

  1. Direct routes from A to C: 3.
  2. Routes through B: 3 roads from A to B × 4 roads from B to C = 12.
  3. Total = 3 + 12 = 15.

Remember · Legs taken one after another multiply; alternative routes add.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2022 · Q55

Medium

There is a numeric lock which has a 3-digit PIN. The PIN contains digits 1 to 7. There is no repetition of digits. The digits in the PIN from left to right are in decreasing order. Any two digits in the PIN differ by at least 2. How many maximum attempts does one need to find out the PIN with certainty?

Answer & explanation

Answer: (c) 10

Each set of three digits from 1–7 that are pairwise at least 2 apart gives exactly one PIN (written in decreasing order). There are 10 such sets, so in the worst case 10 attempts are needed.

  1. Once three digits are chosen, the decreasing order fixes the PIN, so count the digit sets.
  2. The digits must be from 1–7 with every two at least 2 apart. List by the largest digit: 753, 752, 751, 742, 741, 731, 642, 641, 631, 531 — 10 PINs.
  3. Quick count: choosing 3 of 7 numbers with no two consecutive = C(7 − 3 + 1, 3) = C(5, 3) = 10.
  4. In the worst case the right PIN is the last one tried, so up to 10 attempts are needed.

Remember · Choosing k of n numbers with no two consecutive: C(n − k + 1, k).

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2022 · Q76

Easy

One non-zero digit, one vowel and one consonant from English alphabet (in capital) are to be used in forming passwords, such that each password has to start with a vowel and end with a consonant. How many such passwords can be generated?

Answer & explanation

Answer: (c) 945

The password has three characters: the vowel first, the consonant last and the digit in the middle. That gives 5 × 9 × 21 = 945 passwords.

  1. Each password has three characters. It starts with the vowel and ends with the consonant, so the digit sits in the middle.
  2. Choices: 5 vowels (A, E, I, O, U) × 9 non-zero digits × 21 consonants.
  3. Total = 5 × 9 × 21 = 945.

Remember · Fix the forced positions first; each remaining position multiplies by its number of choices.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2022 · Q77

Medium

There are 9 cups placed on a table arranged in equal number of rows and columns out of which 6 cups contain coffee and 3 cups contain tea. In how many ways can they be arranged so that each row should contain at least one cup of coffee?

Answer & explanation

Answer: (d) 81

The 9 cups form a 3 × 3 grid. Choosing the 3 places for tea can be done in C(9, 3) = 84 ways; only the 3 choices that fill a whole row with tea leave a row without coffee. So 84 − 3 = 81.

  1. Equal rows and columns with 9 cups means a 3 × 3 grid.
  2. Cups of the same drink are alike, so an arrangement is fixed by the 3 places holding tea: C(9, 3) = 84.
  3. A row lacks coffee only when all 3 tea cups fill that row: 3 such arrangements.
  4. Arrangements with coffee in every row = 84 − 3 = 81.

Remember · ‘At least one’ counts are easiest as total minus the bad cases.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2022 · Q79

Easy

What is the number of numbers of the form 0·XY, where X and Y are distinct non-zero digits?

Answer & explanation

Answer: (a) 72

X can be any of the 9 non-zero digits and Y any of the other 8, giving 9 × 8 = 72 different decimals.

  1. X can be any of 1–9: 9 choices.
  2. Y must be non-zero and different from X: 8 choices.
  3. Total = 9 × 8 = 72, and each ordered pair gives a different number.

Remember · Distinct choices from the same pool: n × (n − 1) ordered pairs.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

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